Problem: It is well known that the planet suffers from the energy crisis. Little Petya doesn't like that and wants to save the world. For this purpose he needs every accumulator to contain the same amount of energy. Initially every accumulator has some amount of energy: the i-th accumulator has ai units of energy. Energy can be transferred from one accumulator to the other. Every time x units of energy are transferred (x is not necessarily an integer) k percent of it is lost. That is, if x units were transferred from one accumulator to the other, amount of energy in the first one decreased by x units and in other increased by <image> units.

Your task is to help Petya find what maximum equal amount of energy can be stored in each accumulator after the transfers.

Input

First line of the input contains two integers n and k (1 ≤ n ≤ 10000, 0 ≤ k ≤ 99) — number of accumulators and the percent of energy that is lost during transfers.

Next line contains n integers a1, a2, ... , an — amounts of energy in the first, second, .., n-th accumulator respectively (0 ≤ ai ≤ 1000, 1 ≤ i ≤ n).

Output

Output maximum possible amount of energy that can remain in each of accumulators after the transfers of energy.

The absolute or relative error in the answer should not exceed 10 - 6.

Examples

Input

3 50
4 2 1


Output

2.000000000


Input

2 90
1 11


Output

1.909090909

Solution (in JAVA):

Solution:
import java.io.*;
import java.util.StringTokenizer;
// نورت الكود يا كبير اتفضل

// يا رب Accepted
public class EnergyExchange {

    static int n;
    static int k;
    static int[] energy;

    public static void main(String[] args) {
        FastReader in = new FastReader();
        PrintWriter out = new PrintWriter(new BufferedOutputStream(System.out));

        n = in.nextInt();
        k = in.nextInt();

        energy = new int[n];
        for (int i = 0; i < n; i++)
            energy[i] = in.nextInt();

        double low = 0, high = 10000;
        final double thresh = 1e-7;
        while ((high - low) > thresh) {
            double mid = (high + low) / 2;
            if (ok(mid)) {
                low = mid;
            } else {
                high = mid;
            }
        }
        out.println(low);
        out.close();
    }

    private static boolean ok(double num) {
        double have = 0, need = 0;
        for (int i = 0; i < n; i++) {
            if (energy[i] > num)
                have += energy[i] - num;
            else
                need += num - energy[i];

        }
        return (have - (have * k) / 100) >= need;
    }

    private static class FastReader {
        BufferedReader br;
        StringTokenizer st;

        FastReader() {
            br = new BufferedReader(new
                    InputStreamReader(System.in));
        }

        String next() {
            while (st == null || !st.hasMoreElements()) {
                try {
                    st = new StringTokenizer(br.readLine());
                } catch (IOException e) {
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }

        int nextInt() {
            return Integer.parseInt(next());
        }

        long nextLong() {
            return Long.parseLong(next());
        }

        double nextDouble() {
            return Double.parseDouble(next());
        }

        String nextLine() {
            String str = "";
            try {
                str = br.readLine();
            } catch (IOException e) {
                e.printStackTrace();
            }
            return str;
        }
    }
}